A241476 Least positive primitive root g < prime(n) modulo prime(n) with g - 1 a square, or 0 if such a number g does not exist.
1, 2, 2, 5, 2, 2, 5, 2, 5, 2, 17, 2, 17, 5, 5, 2, 2, 2, 2, 65, 5, 37, 2, 26, 5, 2, 5, 2, 10, 5, 65, 2, 5, 2, 2, 82, 5, 2, 5, 2, 2, 2, 101, 5, 2, 170, 2, 5, 2, 10, 5, 26, 37, 26, 5, 5, 2, 26, 5, 26, 5, 2, 5, 17, 10, 2, 37, 10, 2, 2, 5, 26, 10, 2, 2, 5, 2, 5, 17, 26
Offset: 1
Keywords
Examples
a(4) = 5 since 2^2 + 1 = 5 is a primitive root modulo prime(4) = 7, but neither 0^1 + 1 = 1 nor 1^1 + 1 = 2 is a primitive root modulo prime(4) = 7.
Links
- Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
- Z.-W. Sun, New observations on primitive roots modulo primes, arXiv preprint arXiv:1405.0290 [math.NT], 2014.
Programs
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Mathematica
f[k_]:=k^2+1 dv[n_]:=Divisors[n] Do[Do[Do[If[Mod[f[k]^(Part[dv[Prime[n]-1],i]),Prime[n]]==1,Goto[aa]],{i,1,Length[dv[Prime[n]-1]]-1}];Print[n," ",k^2+1];Goto[bb];Label[aa];Continue,{k,0,Sqrt[Prime[n]-2]}];Print[n," ",0];Label[bb];Continue,{n,1,80}] -
PARI
ispr(n,p)=my(f=factor(p-1)[,1],m=Mod(n,p));for(i=1,#f, if(m^(p\f[i])==1, return(0))); m^(p-1)==1 a(n)=my(p=prime(n));for(k=0,sqrtint(p-2),if(ispr(k^2+1,p), return(k^2+1)));0 \\ Charles R Greathouse IV, May 01 2014
Comments